In an L-R-C series circuit, L = 0.280 H and C = 4.00 mF. The voltage amplitude of the source is 120 V. (a) What is the resonance angular frequency of the circuit? (b) When the source operates at the resonance angular frequency, the current amplitude in the circuit is 1.70 A. What is the resistance R of the resistor? (c) At the resonance angular frequency, what are the peak voltages across the inductor, the capacitor, and the resistor?

Z is defined as the impedance of the circuit which is the effective resistance of an electric circuit or component to alternating current, arising from the combined effects of ohmic resistance and reactance.

$\mathbf{Z}\mathbf{=}\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\left(X_L-X_{C)}\right)}^{\mathbf{2}}}$, where Z is the impedance

The equivalent Ohm’s law relation to get the amplitude voltage V in the circuit.

V=IZ

Similarly, the amplitude voltage across the resistor, capacitor and inductor is found by the relation

V=IX

Where X is the reactance which is equal to${\mathbf{X}}_{\mathbf{L}}\mathbf{=}\mathbf{\omega L}$ for inductor and${\mathbf{X}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\omega C}}$ for capacitor and Rfor resistor.

The process of peaking the current at a particular frequency represents the resonance and that frequency is known as resonant frequency.

$\mathbf{\omega }\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{LC}}}$,whereis the resonant angular frequency.

At resonance, ${\mathbf{X}}_{\mathbf{L}}\mathbf{=}{\mathbf{X}}_{\mathbf{C}}$.

At resonance, the impedance is minimum and this minimum value is same as that of resistor.

The maximum voltages across R, C and L is

$$\begin{gathered} {\mathbf{V}}_{\mathbf{R}}\mathbf{=}\mathbf{IR} \\ {\mathbf{V}}_{\mathbf{C}}\mathbf{=}{\mathbf{IX}}_{\mathbf{C}} \\ {\mathbf{V}}_{\mathbf{L}}\mathbf{=}{\mathbf{IX}}_{\mathbf{L}} \end{gathered}$$

$$\begin{gathered} \mathrm{\omega }=\frac{1}{\sqrt{\mathrm{LC}}} \\ \mathrm{L}=0.280\mathrm{mH} \\ \mathrm{C}=4.00\mathrm{\mu F} \end{gathered}$$

$$\begin{gathered} \mathrm{\omega }=\frac{1}{\sqrt{\left(0.280\mathrm{H}\right)\left(4\times {10}^{-6}\mathrm{C}\right)}} \\ =945\frac{\mathrm{rad}}{\mathrm{s}} \end{gathered}$$

Therefore, the resonance angular frequency of the circuit is $945\frac{\mathrm{rad}}{\mathrm{s}}$.

At resonance,

$Z=R$

$\mathrm{Z}=\frac{\mathrm{V}}{\mathrm{I}}$

$$\begin{gathered} \mathrm{I}=1.70\mathrm{A} \\ \mathrm{V}=120\mathrm{V} \\ \mathrm{Z}=\frac{120\mathrm{V}}{1.70\mathrm{A}} \\ =70.6\mathrm{\Omega } \\ \mathrm{Z}=\mathrm{R} \\ \mathrm{R}=70.6\mathrm{\Omega } \end{gathered}$$

Therefore, Resistance R of the resistor .$70.6\Omega$

$$\begin{gathered} {\mathrm{V}}_{\mathrm{R}}=120\mathrm{V} \\ {\mathrm{V}}_{\mathrm{C}}=\frac{1}{\mathrm{\omega C}} \\ =\frac{1.70\mathrm{A}}{\left(945{\displaystyle \frac{\mathrm{rad}}{\mathrm{s}}}\right)\left(4\times {10}^{-6}\mathrm{F}\right)} \\ =450\mathrm{V} \\ {\mathrm{V}}_{\mathrm{L}}=\mathrm{I\omega L} \\ =\left(1.70\mathrm{A}\right)\left(945\frac{\mathrm{rad}}{\mathrm{s}}\right)\left(0.280\mathrm{H}\right) \\ =450\mathrm{V} \end{gathered}$$

At resonance,${\mathit{V}}_{\mathbf{R}}$ is the same as the source${\mathit{V}}_{\mathbf{L}}$ and${\mathit{V}}_{\mathbf{C}}$ are the same, and${\mathit{V}}_{\mathbf{L}}\mathbf{-}{\mathit{V}}_{\mathbf{C}}$ =0.

Therefore, a)The resonance angular frequency of the circuit is $945\frac{\mathrm{rad}}{\mathrm{s}}$.b) Resistance R of the resistor $70.6\Omega$.c)The peak voltages across the inductor is 450V, the capacitor is and the resistor is 120V.